Question: Solve for $x$ : $2x^2 + 20x + 32 = 0$
Dividing both sides by $2$ gives: $ x^2 + {10}x + {16} = 0 $ The coefficient on the $x$ term is $10$ and the constant term is $16$ , so we need to find two numbers that add up to $10$ and multiply to $16$ The two numbers $2$ and $8$ satisfy both conditions: $ {2} + {8} = {10} $ $ {2} \times {8} = {16} $ $(x + {2}) (x + {8}) = 0$ Since the following equation is true we know that one or both quantities must equal zero. $(x + 2) (x + 8) = 0$ $x + 2 = 0$ or $x + 8 = 0$ Thus, $x = -2$ and $x = -8$ are the solutions.